Yes the product $\nu \lambda$ makes sense as a velocity. Defining $E = \hslash \omega$ and $p=\hslash k$ (the Planck constant $h=2\pi \hslash$, where the $2\pi$ is injected into the $\hslash$, since physicists usually prefer to discuss the angular frequency $\omega=2\pi\nu$ and the wave vector $k=2\pi p$ rather than the frequency $\nu$ and the momentum $p$ for Fourier-transform notational convenience), you end up with
$$\nu \lambda = \frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$$
which is the standard definition for the phase velocity. It corresponds to the velocity of the wave component at frequency $\nu$ propagating over distance $\lambda$ per unit time.
This is always true, but for more complicated situation, a system is represented by a superposition of different waves propagating at different phase velocity. It results a wave-packet propagating, as an ensemble, at the group velocity
$$v_{g}=\frac{\partial \omega \left(k\right)}{\partial k}$$
where the dispersion relation of the wave-packet is noted $\omega \left(k\right)$.
Only the group velocity has some physical clear interpretations. For instance, the phase velocity can be larger than the speed of light, but the group velocity can never been larger than the speed of light $v_{g}\leq c$, at least in vacuum.
For a photon in free space for instance, $\omega \left(k\right)=c k$ and thus it's group velocity is $c$. For a free non-relativistic particle of mass $m$, $\omega = \hslash k^2/2m$, and $v_{g}=\hslash k/m$. etc...
For a human body, you have to count all the atoms constituting the body. The individual frequency of one atom interferes with the frequencies of all the others, resulting in an almost flat dispersion relation. The group velocity is then ridiculously small. A human body does not move due to quantum effect ! You can get a basic idea of the group velocity of a human body supposing you are a free particle of mass $100 \textrm{kg}$ and wavelength $1 \textrm{m}^{-1}$ (i.e. the order of magnitude of your size is about $1 \textrm{m}$) the $\hslash$ kills $v_{g} \sim \left(10^{-36}-10^{-34}\right) \textrm{m.s}^{-1}$ !
More about that :
1) Yes. The photon's matter wave is actually its electromagnetic wave.
2) Photon emission is not such a kind of process when you get some particles emitted in some interval of time, and you can assume some emission moment within that interval to every particular particle. No, it is a quantum process. The system's intermediate states are not "some particles are emitted, and some are not emitted yet", but "system has some intemediate probability to be in the initial state, and some probability to be in the final state". Every photon is the subject to that accumulating probability. So the most we can say for every photon is that it is emitted in the same interval of time.
The frequency, polarisaion, direction, spatial distribution and all such characteristics of each photon would be the same as those of the electromagnetic wave. (Some advanced details are omitted.)
3) The same way as you may interpret some particular waveform in different bases, like a function of time or a set of sinewave weights, you can interpret photons in different bases. The non-basic waveform is then understood as a quantum superposition of states that belong to the chosen basis.
4) The question is based on the wrong assumption, see the answer to the question 2.
P.S. You deleted your 4 questions, leaving just one, but I hope my answers would still help you.
Best Answer
The wavelength $\lambda$ has a direction. Or, more precisely, the wave number $$\vec{k}=2\pi \begin{pmatrix}1/\lambda_x \\ 1/\lambda_y \\ 1/\lambda_z \end{pmatrix}$$ is a vector quantity (with a direction perpendicular to the wave fronts).
(image from my answer to Significance of wave number?)
Therefore it makes more sense to relate the wave number $\vec{k}$ via $$\vec{p}=\frac{h}{2\pi}\vec{k}$$ to the momentum $\vec{p}$ which is also a vector quantity, and not to the energy $E$ which is a scalar quantity. It also fits nicely into special relativity where $(E/c, \vec{p})$ make up a 4-vector and $(\omega/c, \vec{k})$ make up another 4-vector. The four-vector $(\omega/c,\vec{k})$ is established by its 4-product with the 4-vector $(ct,\vec{x})$ $$\phi=-\omega t+\vec{k}\cdot\vec{x}$$ giving the phase which is a 4-scalar.
But nevertheless, in 1924 when de Broglie brought up this hypothesis, even this was still a speculative guess. Only later it was confirmed experimentally that this relation is actually true, not only for massless photons, but also for massive particles. The first experiment of this kind was the Davisson-Germer experiment (1923-1927). It involved electrons (with known momentum $\vec{p}=m\vec{v}$) scattered by the surface of a nickel crystal (with known atomic grid distance $d$). From the observed diffraction pattern and the atomic grid distance $d$ they could calculate the wavelength $\lambda$ of the electrons, and found it matched the wavelength predicted by de Broglie's $\lambda=h/p$.